Question #d4b06

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Answer 1 · 2016-09-22T18:22:23

$Reqd. Lim.= π .$ $" Reqd. Lim.="pi.$

Let $\frac{π}{x} = θ, so that, as x \to \infty, θ \to 0 .$ $pi/x=theta", so that, as "xrarroo, thetararr0.$

Also, $x = \frac{π}{θ}$ $x=pi/theta$

We know that, $lim_(yrarr0)siny/y=1.$

Now, reqd. Lim. $=lim_(thetararr0) pi/theta*sintheta$

$=pi{lim_(thetararr0)sintheta/theta}$

$=pi.$

Enjoy Maths.!